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Question 4 Iowa Families Question1 For a random sample of 64 Iowa homes, average weekly food...

Question 4 Iowa Families Question1 For a random sample of 64 Iowa homes, average weekly food expenditure turns out to be $160, with a standard devaition of $64. Let mu denote the mean weekly food expenditure for Iowa families. Find a 95% confidence Interval for μ.

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Answer #1

4)

Solution :

t /2,df = 1.998

Margin of error = E = t/2,df * (s /n)

= 1.998 * (64 / 64)

Margin of error = E = 16.01

The 95% confidence interval estimate of the population mean is,

- E < < + E

160 - 16.01 < < 160 + 16.01

144.01 < < 175.99

(144.01 , 175.99)

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